How does mass affect slipping? I experimented by rolling a cylinder down a tube and noticed that by increasing its mass, it's time taken to reach the bottom decreased. However, mass is independent from the time taken. The cylinder is dropped from the top of a ramp through the starting light gate. 
Thus, there must be an effect of slipping along with the cylinder's rotational kinetic energy to change the amount of time. Therefore, in this situation, how does changing the center of mass affect its tendency to slip? 
 A: There are mathematical ways of showing all this, but it is important to have intuition before starting to calculate. Here is my intuitive picture:
The kinetic energy of the rolling cylinder has two components:


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*The kinetic energy of its motion along the plane.

*The kinetic energy of its rotation about its own axis.
The only source of kinetic energy is the loss of potential energy from being lower down the plane than at the start. So at a given point along the plane, the total kinetic energy is constant.
Thus 1, the kinetic energy of motion along the plane, depends on 2, the kinetic energy of rotation. The more kinetic energy of rotation there is, the less kinetic energy of motion there is, and therefore the slower the speed of the cylinder down the plane.


*

*If the cylinder had all its mass at its exact centre, its kinetic energy of rotation would be zero, and so its kinetic energy of motion along the plane would be at a maximum - and so would its speed.

*If the cylinder has all its mass on its periphery, its kinetic energy of rotation would be much higher, and so its kinetic energy of motion along the plane would be much lower - and so would its speed.
This is why the speed depends on the distribution of mass in the cylinder.
A: Thank you for the update, WJ47.
The slope of the blue tube looks very steep.  Both the ruler/tube and the white cylinder (cellotape holder) look quite smooth, so I think there will be little friction, resulting in a mixture of rolling and sliding here.  It is very difficult to predict how much of each.  This is a rather 'messy' experiment, IMHO, difficult to tie up with theory. 
Another difficulty is the timing.  It seems you are timing the cylinder between two points at which it has unknown speed.  You can calculate the average speed from start to finish, but this does not tell you the acceleration. 
I recommend that you change the experimental apparatus and do the experiment again.  Allow the white cylinder to roll down a wooden incline which is much less steep (and not so smooth) - no more than about $30^{\circ}$.  (Check that it does roll not slide.)  This will make the time of descent longer, so results should be more accurate. I would also weigh the white cylinder before each run.
Make sure the putty is evenly spread in the hole - that way you can calculate moment of inertia and use the following theory.  Start timing as soon as the cylinder begins to move - then you can assume the initial speed is zero.  
Theory
If the cylinder rolls down the incline without sliding/slipping then there is no loss of energy due to friction. Initial PE + KE = final PE + KE.  Assuming it starts from rest and falls through vertical height $h$ then
$$\frac12Mv^2+\frac12I\omega^2 = Mgh$$
where $M$ is mass, $v$ is final velocity of the centre of mass CM, $\omega=\frac{v}{R}$ is final angular velocity, $R$ is outer radius of cylinder, and $I=kMR^2$ is its moment of inertia.  $k$ is a variable related to the distribution of mass in the cylinder.  Putting these into the formula and rearranging we get
$$(1+k)v^2 = 2gh$$
Using the kinematic equations for constant acceleration, distance $L$, time $t$ and final velocity $v$ down the plane are related by 
$$L = average velocity \times time = \frac12(u+v)t = \frac12 vt$$
This works because I am assuming the initial velocity $u=0$.  Substituting in the above equation and rearranging we get
$$t^2 = 2(1+k)\frac{L^2}{gh}$$
You should get a straight line if you plot $t^2$ against $k$.  The only remaining difficulty is to calculate the value of $k$ each time you add mass to the cylinder. 
Suppose the sellotape holder has mass $m_1$ and outer & inner radii $R$ and $r_1$. Suppose you add mass by fixing a ring of plasticine of mass $m_2$ inside the holder, leaving a central hole of radius $r_2$.  
The moment of inertia (MI) about the centre for the holder is $I_1 = \frac12 m_1(R^2+r^2)$, and for the added plasticine is $I_2=\frac12 m_2 (r_1^2+r_2^2)$.  The MI of the whole is
$$I = I_1 + I_2 = \frac12(m_1R^2+m_1r_1^2+m_2 r_1^2+m_2 r_2^2) = \frac12(\mu_1+\rho_1^2+\mu_2\rho_2^2)MR^2$$
so that
$$k = \frac12(\mu_1+\rho_1^2+\mu_2\rho_2^2)$$
where $\mu_1=\frac{m_1}{M}$, $\mu_2=\frac{m_2}{M}$, $\rho_1=\frac{r_1}{R}$, $\rho_2=\frac{r_2}{R}$ and $M=m_1+m_2$.
You will need to re-weigh the mass $M$ of the holder & plasticine each time you add more, also re-measure the inner radius $r_2$, and re-calculate $k$ using the above formulas.
Because $\mu_1$ and $\rho_1$ do not change, you could instead calculate $p=\mu_2 \rho_2^2$ and plot $t^2$ against $p$ to get a straight line.  $p=0$ when the cellotape holder is empty.
